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Since the volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.If the volume were measured by displacement of such water, there would be no effect on the density of the solid. But the measurement would be inaccurate, leading to a misrepresentation of the actual density.

Density = (mass)/(volume)

Assume that the mass had been accurately measured by other means, and the volume then measured by displacement of water that included air bubbles.

During the overflow procedure, the air would escape, and only the H2O component of the displaced fluid would be available for measurement. Hence, the volume of the solid would be under-reported.

Since (volume) is the denominator of the fraction, the apparently smaller volume would cause the quantity (mass)/(volume) to become artificially large, and the density of the solid would appear larger than its actual density.

--relxerd

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13y ago
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9y ago

In the water displacement method, the object MUST be completely submerged in the water. If it isn't, only the volume of the submerged portion will displace the water and the portion floating above the level of water can not be measured.

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9y ago

If there are air bubbles present in the water during a water displacement, the density of a solid is lower than its actual density. This is because the air bubbles will tend to increase the measured volume. Since density and volume are indirectly proportional, an increase in volume will decrease the density.

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10y ago

Air bubbles don't affect the density of a solid. But if there are air bubbles

in a sample of the solid, then its density APPEARS to be less, because in

some of the places where you think there's solid, there's only air.

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14y ago

If the volume were measured by displacement of such water, there would be no effect on the density of the solid. But the measurement would be inaccurate, leading to a misrepresentation of the actual density. Density = (mass)/(volume) Assume that the mass had been accurately measured by other means, and the volume then measured by displacement of water that included air bubbles. During the overflow procedure, the air would escape, and only the H2O component of the displaced fluid would be available for measurement. Hence, the volume of the solid would be under-reported. Since (volume) is the denominator of the fraction, the apparently smaller volume would cause the quantity (mass)/(volume) to become artificially large, and the density of the solid would appear larger than its actual density.

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16y ago

It would lighten the density, but the air would float to the top and disperse.

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14y ago

What would be the effect on the density of a solid whose volume was determined by water displacement if air bubbles were present in the water?

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14y ago

Properly calculated, the density should incorporate the bubbles. If the bubbles were to "escape", the density would increase and need to be recalculated to be accurate.

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15y ago

dont cheat

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Q: How does air bubbles affect density of a solid?
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